A long solenoid has `500` turns. When a current of `2A` is passed through it, the resulting magnetic flux linked with each turn of the solenoid is `4xx10^(-3)Wb` . The self-inductance of the solenoid is

Options

A. 2 H
B. 1.5 H
C. 1 H
D. 0.5 H

Answer

C

Explanation

The total magnetic flux `(phi)` linked with the solenoid is given by `phi = 500 xx 4 xx 10^(-3) = 2 Wb`
But for a coil `phi = Li :. L = (phi)/(i) = (2Wb)/(2A) = 1H`.